Math1502FinalExamSp2010

Math1502FinalExamSp2010 - MATH1502 G Group Calculus II...

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MATH1502 - G Group Calculus II Final Exam Corrected Wednesday May 5th, 2010, 2:50 - 5:40 No calculators or notes are allowed. There are 140 marks on the °nal exam, 130=100%. Name_______________________________________ Student Number ______________________________ Group G ___________________________________ TA _________________________ ______________ Question Grade Out of 1 14 2 15 3 24 4 40 5 47 Total 140 1
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Question 1 (a) Find the radius of convergence and interval of convergence of the power series 1 X k =2 1 ° 1 3 cos ( k + 1) 7 = 5 (14 x ) k : (8 marks) Solution We use the root test applied to a k = ° ° ° ° ° 1 ° 1 3 cos ( k + 1) 7 = 5 (14 x ) k ° ° ° ° ° = 1 ° 1 3 cos ( k + 1) 7 = 5 (14 j x j ) k : (Recall j cos j ± 1 ). We see that lim k !1 a 1 =k k = lim k !1 1 ° 1 3 cos ( k + 1) 7 = 5 (14 j x j ) k ! 1 =k = lim k !1 1 ° 1 3 cos ( k + 1) 7 = 5 ! 1 =k (14 j x j ) = 14 j x j : (2 marks) ( Justi°cation of this, not really required from students: Recall lim k !1 k 1 =k = 1 , so 2 3 1 k 7 = 5 ± 1 + 1 k ² 7 = 5 ± 1 ° 1 3 cos ( k + 1) 7 = 5 ± 4 3 1 k 7 = 5 : and then ³ 2 3 ´ 1 =k 1 ( k 1 =k ) 7 = 5 ± 1 + 1 k ² 7 = (5 k ) ± 1 ° 1 3 cos ( k + 1) 7 = 5 ! 1 =k ± ³ 4 3 ´ 1 =k 1 ( k 1 =k ) 7 = 5 : By the pinching/ sandwich rule, and as lim k !1 ± 4 3 ² 1 =k = 1 and lim k !1 ± 1 + 1 k ² 1 =k = 1 ; lim k !1 1 ° 1 3 cos ( k + 1) 7 = 5 ! 1 =k = 1 : ) 2
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By the root test, we have convergence for j x j < 1 14 : And we have divergence for j x j > 1 . So the radius of convergence is 1 14 . (2 marks) Next, we test the endpoints x = ² 1 14 . x = ° 1 14 Here 1 X k =2 1 ° 1 3 cos ( k + 1) 7 = 5 (14 x ) k = 1 X k =2 1 ° 1 3 cos ( k + 1) 7 = 5 ( ° 1) k : Now taking absolute values, ° ° ° ° ° 1 ° 1 3 cos ( k + 1) 7 = 5 ( ° 1) k ° ° ° ° ° ± 4 3 k 7 = 5 and note that the series P 1 k 7 = 5 converges ( p ° series with p = 7 5 > 1 ). The comparison test shows that 1 X k =2 1 ° 1 3 cos ( k + 1) 7 = 5 ( ° 1) k converges absolutely, and hence converges. x = 1 14 Here again 1 X k =2 1 ° 1 3 cos ( k + 1) 7 = 5 (14 x ) k = 1 X k =2 1 ° 1 3 cos ( k + 1) 7 = 5 and the argument used above shows that the series converges absolutely, and hence converges. So the interval of convergence is µ ° 1 14 ; 1 14 : (4 marks) 3
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(b) Find the radius of convergence of the power series 1 X k =0 ( k !) 2 (3 k ) 2 k x k : (6 marks) You may assume the limit lim k !1 ³ 1 + 1 k ´ k = e: Solution Because of the factorials, we use the ratio test, applied to a k = ° ° ° ° ° ( k !) 2 (3 k ) 2 k x k ° ° ° ° ° = ( k !) 2 (3 k ) 2 k j x j k ; with x 6 = 0 . We see that lim k !1 a k +1 a k = lim k !1 (( k +1)!) 2 (3( k +1)) 2 k +2 · j x j k +1 ¸ ( k !) 2 (3 k ) 2 k j x j k (2 marks) = lim k !1 k 2 k ( k + 1) 2 k +2 3 2 k 3 2 k +2 ³ ( k + 1)! k ! ´ 2 j x j = lim k !1 1 ± 1 + 1 k ² 2 k ( k + 1) 2 1 3 2 ( k + 1) 2 j x j = 1 e 2 3 2 j x j : (2 marks) By the ratio test, this converges for 1 e 2 3 2 j x j < 1 , j x j < 9 e 2 : (2 marks) Moreover, it diverges for j x j > 9 e 2 . So the radius of convergence is 9 e 2 . 4
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Question 2 (a) Find the solution of the di/erential equation y 0 cos x + 1 + y = 4 that satis°es y ( ° ) = 0 : (8 marks) Solution We want the form y 0 + p ( x ) y = q ( x ) ; so multiply by cos x + 1 : y 0 + (cos x + 1) y = 4 (cos x + 1) : (1 mark) Then p ( x ) = cos x + 1 , and our integrating factor is e ( R (cos x +1) dx ) = e sin x + x : (2 marks) Multiply by e sin x + x : e sin x + x y 0 + e sin x + x (cos x + 1) y = 4 e sin x + x (cos x + 1) ) d dx ± e sin x + x y ² = 4 e sin x + x (cos x + 1) : We integrate this to obtain e sin x + x y = 4 Z e sin x + x (cos x + 1) dx + C = 4 e sin x + x + C: Then
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